Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using V = vector<T>;
int R, C;
int G[100][100];
int cnt = 0;
stringstream ans;
bool solve() {
  int sum = 0;
  for (int(r) = 0; (r) < (R); ++(r)) {
    for (int(c) = 0; (c) < (C); ++(c)) {
      sum += G[r][c];
    }
  }
  if (!sum) return 1;
  ++cnt;
  int row = -1;
  for (int(r) = 0; (r) < (R); ++(r)) {
    bool g = 1;
    for (int(c) = 0; (c) < (C); ++(c)) {
      if (!G[r][c]) {
        g = 0;
        break;
      }
    }
    if (g) {
      row = r;
      break;
    }
  }
  int col = -1;
  for (int(c) = 0; (c) < (C); ++(c)) {
    bool g = 1;
    for (int(r) = 0; (r) < (R); ++(r)) {
      if (!G[r][c]) {
        g = 0;
        break;
      }
    }
    if (g) {
      col = c;
      break;
    }
  }
  if (row != -1 && col != -1) {
    if (C > R) {
      ans << "row " << row + 1 << "\n";
      for (int(c) = 0; (c) < (C); ++(c)) {
        --G[row][c];
      }
      return solve();
    } else {
      ans << "col " << col + 1 << "\n";
      for (int(r) = 0; (r) < (R); ++(r)) {
        --G[r][col];
      }
      return solve();
    }
  } else if (row != -1) {
    ans << "row " << row + 1 << "\n";
    for (int(c) = 0; (c) < (C); ++(c)) {
      --G[row][c];
    }
    return solve();
  } else if (col != -1) {
    ans << "col " << col + 1 << "\n";
    for (int(r) = 0; (r) < (R); ++(r)) {
      --G[r][col];
    }
    return solve();
  }
  return 0;
}
void test() {
  default_random_engine re;
  for (int tc = 0; tc < 100000; ++tc) {
    R = re() % 5 + 1;
    C = re() % 5 + 1;
    int stuff = re() % 5;
    for (int(i) = 0; (i) < (stuff); ++(i)) {
      if (re() % 2) {
        int row = re() % R;
        for (int(c) = 0; (c) < (C); ++(c)) {
          G[row][c]++;
        }
      } else {
        int col = re() % C;
        for (int(r) = 0; (r) < (R); ++(r)) {
          G[r][col]++;
        }
      }
    }
    cout << R << " " << C << endl;
    for (int(r) = 0; (r) < (R); ++(r)) {
      for (int(c) = 0; (c) < (C); ++(c)) {
        cout << G[r][c] << " ";
      }
      cout << endl;
    }
    cnt = 0;
    assert(solve());
    cout << cnt << " " << stuff << endl;
    assert(cnt <= stuff);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> R >> C;
  for (int(r) = 0; (r) < (R); ++(r)) {
    for (int(c) = 0; (c) < (C); ++(c)) {
      cin >> G[r][c];
    }
  }
  if (!solve()) {
    cout << -1 << endl;
  } else {
    cout << cnt << endl << ans.str();
  }
}